# How can the SI unit of resistivity [Ωm] be interpreted?

The way I interpret the resistance $R$ of a resistor, which has dimensions $[\frac{\mathrm{V}}{\mathrm{A}}]$ is: how many volts must be applied across the resistor to achieve 1 ampere of current?

The conductance $G$, which has dimensions $[\frac{\mathrm{A}}{\mathrm{V}}]$ is then: how many amperes of current flow through the resistor when applying 1 volt?

I realize that these quantities are related to the geometry, whereas the resistivity $\rho = R\times \frac{A}{L}$ which has dimensions $[\Omega \cdot \mathrm{m}]$ is an intrinsic property of the material (doping of semi-conductor, electron/hole mobility, etc).

However, I cannot achieve an intuitive understanding to interpret the dimensions of resistivity. Can this be clarified?

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$\Omega$m is the simplified unit of resistivity. The full unit is $\Omega$m$^2/$m. This means that a given length of material with a given cross sectional area will have a certain resistance whose value can be calculated using the resistivity.

For a 1 m length of material with a 1 mm$^2$ cross sectional area and a resistivity of 1:

$1 \Omega \mathrm{m} = R(10^{-6}\mathrm{m}^2/1\mathrm{m})$

$R = {1\Omega \mathrm{m} \over(10^{-6}\mathrm{m}^2/1\mathrm{m})} = 10^6\Omega$

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Maybe it's easier to understand intuitively if you don't reduce the dimensions, in the same way of thinking as the units of gain as volts/volts, you could think of the units of resistivity as $\frac {\Omega \mathrm{m}^2}{\mathrm{m}}$, which fits with the physical interpretation of a resistive object of constant cross-sectional area and a given length.

Consider also the usual dimension for sheet resistivity, which is $\Omega$ per square, where "per square" really doesn't mean anything dimensionally, but prevents confusion with simply saying $\Omega$.

Another example, the units of torque ($\mathrm{n}\cdot \mathrm{m}$) are the same as the units of work- it's the physical interpretation that makes the difference.

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The simplified unit Ω·m can be interpreted as:

If you multiply it by a length — specifically a thickness — then you obtain the resistance of an arbitrarily-sized square of that material with that thickness.

This is less useful in practical situations than the un-simplified interpretation.

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